浙江财经大学
信工学院ACM集训队

HDU 1217 Arbitrage(Floyd)

本文由 Ocrosoft 于 2016-08-06 16:12:24 发表

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7024    Accepted Submission(s): 3227

Problem Description
Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent. 
Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.
 

Input
The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n. 
 

Output
For each test case, print one line telling whether arbitrage is possible or not in the format “Case case: Yes” respectively “Case case: No”. 
 

Sample Input
3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0
 

Sample Output
Case 1: Yes
Case 2: No
 

Solution

题意:给出一些货币,然后是一些货币之间的汇率,汇率不能反向计算。问能不能通过买其他货币来获利。

不能反向计算所以是有向边;刚开始edge[i][i]=1,如果结束之后edge[i][i]>1,说明获利了。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 30+10
using namespace std;
double edge[MAXN][MAXN];
int find(map<string,int> mp,string s)
{
    if(mp.count(s))return mp[s];
    else return -1;
}
int main()
{
    int n,m,kase=0;
    string s,t;
    double r;
    map<string,int> mp;
    while(cin>>n&&n)
    {
        kase++;
        for(int i=1; i<=n; i++) //初始化
        {
            string s;
            cin>>s;
            mp[s]=i;
            for(int j=1;j<=n;j++)edge[i][j]=(i==j?1:INF);
        }
        cin>>m;
        for(int i=1; i<=m; i++)
        {
            cin>>s>>r>>t;
            int a=find(mp,s),b=find(mp,t);
            if(edge[a][b]==INF)edge[a][b]=r;
            else edge[a][b]=max(edge[a][b],r);
            //if(edge[b][a]==INF)edge[b][a]=1/r;
            //else edge[b][a]=max(edge[b][a],1/r);
        }
        //Floyd
        for(int k=1; k<=n; k++)
            for(int i=1; i<=n; i++)
                for(int j=1; j<=n; j++)
                    if(edge[i][k]<INF&&edge[k][j]<INF)
                        edge[i][j]=(edge[i][j]==INF?edge[i][k]*edge[k][j]:max(edge[i][j],edge[i][k]*edge[k][j]));
        bool flag=0;
        for(int i=1;i<=n;i++)
            if(edge[i][i]>1){flag=1;break;}
        printf("Case %d: %s\n",kase,flag?"Yes":"No");
    }
    return 0;
}

 

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